在△ABC中,∠C=90°,AC=BC=根号2+1,AD是∠CAB的角平分线,求CD的长

过D作DE⊥AB
∵∠C=90°,AC=BC=√2+1
∴AB²＝AC²+BC²＝2（√2+1）²
∴AB＝√[2（√2+1）²]＝（√2+1）×√2
∵AD平分∠CAB,∠C=90°,DE⊥AB
∴CD＝DE
∴S△ACD＝AC×CD/2
S△ADB＝AB×DE/2＝AB×CD/2
S△ABC＝AC×BC/2
∵S△ABC＝S△ACD+ S△ADB
∴AC×CD/2+AB×CD/2＝AC×BC/2
AC×CD+ AB×CD＝AC×BC
（AC+AB）×CD＝AC×BC
[√2+1+（√2+1）×√2] ×CD＝（√2+1）²
（√2+1）²×CD＝（√2+1）²
CD＝1

tan(A/2)=(1-cosA)/sinA
Rt△ACD,tan∠CAD=CD/AC
CD=tan∠CAD * AC=tan(π/8) *AC
tan(π/8)=(1-cosπ/4)/sin(π/4)=√2-1
CD=(√2-1)*(√2+1)=1内个 tan cos 是什么意思你还没学三角函数？我只有初二 还没有学到那好就这么解答吧，过D作DE垂直AB于E ...